A206922 Root of the n-th binary palindrome. Least number r > 1 such that A006995(n) can be represented by a finite or infinite number of iterations A006995(A006995(A006995(...(...(r))...).

2, 2, 3, 4, 4, 6, 4, 8, 6, 10, 11, 12, 13, 14, 4, 16, 8, 18, 19, 20, 6, 22, 23, 24, 25, 26, 10, 28, 29, 30, 11, 32, 12, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 13, 46, 47, 48, 49, 50, 14, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 4, 64, 16, 66, 67, 68

Offset

1,1

Comments

If n is not a binary palindrome, then a(n)=n.

For n>3: a(n)<n, iff n is a binary palindrome.

For n<>3: The number of iterations such that A006995(n)= A006995(A006995(A006995(...(...(r))...) is given by A206921(n).

Links

Table of n, a(n) for n=1..68.

Formula

a(n) <= n for n > 1.

a(n)=p(k), where p(k) can be determined by the following iteration: set k=0, p(0)=A006995(n). Repeat while A178225(p(k))==1, set k=k+1, p(k)=A206915(p(k-1)) end repeat [for n<>3].

Recursion for n<>3:

Case 1: a(n)=n, if n is not a binary palindrome;

Case 2: a(n)=a(A206915(n)), else.

Formally: a(n)=if (A178225(n)==0) then n else a(A206915(n)).

Example

a(1)=2, since A006995(1) = 0 = A006995(A006995(2)).
a(3)=3, since A006995(3) = 3 = A006995(A006995(A006995(...(3)...).
a(7)=4, since A006995(7) = 15 = A006995(A006995(A006995(4)).
a(9)=6, since A006995(9) = 21 = A006995(A006995(6)).

Prog

/* C program fragment, omitting formal details, n!=3 */
k=0;
p=A006995(n);
while A178225(p)==1
{
  k++;
  p=A206915(p);
}
return p;

Crossrefs

Cf. A006995, A206921, A178225, A206915, A154809.

Sequence in context: A289677 A113967 A205386 * A276775 A271169 A352745

Adjacent sequences: A206919 A206920 A206921 * A206923 A206924 A206925

Keyword

nonn,base

Author

Hieronymus Fischer, Mar 12 2012

Status

approved